Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour. On the return trip, she travels the same distance in 3/5 of an hour. What is the average rate of speed of the wind and the average rate of speed of the plane?
Answers
the average airspeed of the plane = 42 miles per hour
the average wind speed = 12 miles per hour
Let x be the average airspeed of the plane.
Let y be the average wind speed.
Distance =time * speed
Initial trip: 18 = \frac{1}{3}(x+y)18=
3
1
(x+y)
Return trip: 18 = \frac{3}{5}(x+y)18=
5
3
(x+y)
We solve for x and y
18 = \frac{1}{3}(x+y)18=
3
1
(x+y)
Multiply both sides by 3
54= x+ y
y= 54- x ------------> equation 1
18 = \frac{3}{5}(x+y)18=
5
3
(x+y)
Multiply both side by 5
90 = 3(x-y)
90= 3x- 3y ------------------> equation 2
Plug in y=54-x in second equation
90= 3x- 3(54-x)
90 = 3x - 162 + 3x
90 = -162 + 6x
Add 162 on both sides
252= 6x
Divide both sides by 6
So x= 42
y= 54- x
Plug in 42 for x
y= 54 - 42= 12
the average airspeed of the plane = 42 miles per hour
the average wind speed = 12 miles per hour
Given:
The tailwind to the nearby town in 1/3 of an hour.
On a return trip, the same distance is covered in 3/5 hours of an hour.
To Find:
The average speed of the wind and the average rate of speed of the plane.
Solution:
Let the average speed of the plane be x
Let the average wind speed be y
The formula to calculate distance is time × speed
So, the initial trip: 18 = 1/3(x + y)
Return trip :18 = 3/5(x + y)
Now we solve x and y.
18 = 1/3(x + y)
Multiplying both the sides by 3
So, 54 = x+ y
y = 54-x ..(i)
Then, 18 = 3/5(x + y)
Again multiplying both the sides by 5
90 = 3(x-y)
90 = 3x - 3y ..(ii)
Now, substituting the value of y in equation (ii)
90 = 3x - 3(54-x)
90 = 3x - 162 + 3x
90 = -162 + 6x
Adding 162 on both the sides
252 = 6x
Now, dividing both the sides by 6
So, x = 42
y = 54-x
Substituting the value of x
y = 54 - 42
y = 12
So, the average airspeed of the plane = 42 miles per hour and the average wind speed = 12 miles per hour.